= Solution
Choose any nonzero vector $b$ in the one-dimensional <kernel> of $A_k$. It gives $\sum_{j=1}^ka_{ij}b_j=0$ for every row. Because no column is zero, $b$ must have at least two nonzero entries: a relation supported at a single index would make that column zero.
Suppose, contrary to the claim, that $A_k\operatorname{diag}(\lambda_1,\ldots,\lambda_k)b=0$. Its <kernel> is spanned by $b$, so there is a scalar $c$ with $\lambda_jb_j=cb_j$ for every $j$. At the two nonzero entries this gives two equal weights $\lambda_j=c$, contradicting their distinctness. Thus \b[the weighted sum is nonzero in at least one row]. This proves that <distinct diagonal weights break a first column dependence>. Some entries of $b$ may be zero; only two nonzero entries are needed.
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